Series Convergence of Harmonic Means

Question

Let $\{x_n\}$ be a sequence of real numbers such that $0< x_1 <x_2$. If $$x_n= \frac{2}{\frac{1}{x_{n-1}}+\frac{1}{x_{n-2}}}, $$then show that $$\lim_{n\to\infty}x_n=\frac{3x_1x_2}{2x_1+x_2}.$$

Answer 1

Note that \begin{align} & x_n = \frac{2}{\frac{1}{x_{n-1}} + \frac{1}{x_{n-2}}} \\ \iff & \frac{1}{x_{n-1}} + \frac{1}{x_{n-2}} = \frac{2}{x_{n}} \\ \iff & a_n = \frac{1}{2}a_{n-1} + \frac{1}{2}a_{n-2}\\ \iff & a_n-a_{n-1} = -\frac{1}{2}\left(a_{n-1}-a_{n-2}\right) \end{align} where $a_i = \frac{1}{x_i}$. Consequently, $$\sum_{n=3}^{k}(a_n-a_{n-1}) = a_k-a_2 = -\frac{1}{2}\sum_{n=3}^{k}(a_{n-1}-a_{n-2}) = -\frac{1}{2}(a_{k-1}-a_1)$$ Let $\alpha = \lim_{k\to \infty} a_k$, then the above equation implies $$\alpha -a_2 =-\frac{1}{2}(\alpha-a_1) \implies \alpha = \frac{2a_2+a_1}{3}.$$ Therefore, $$\lim_{n\to\infty}x_n = \frac{1}{\alpha} = \frac{3x_1 x_2}{2x_1 + x_2}.$$

Answer 2

Note that \begin{align} & x_n = \frac{2}{\frac{1}{x_{n-1}} + \frac{1}{x_{n-2}}} \\ \iff & \frac{1}{x_{n-1}} + \frac{1}{x_{n-2}} = \frac{2}{x_{n}} \\ \iff & a_n = \frac{1}{2}a_{n-1} + \frac{1}{2}a_{n-2}, \end{align} where $a_i = \frac{1}{x_i}$.

It's easy to verify that $$a_n = \alpha + \beta\left(-\frac{1}{2}\right)^n.$$ Consequently, $$\lim_{n\to \infty}a_n = \alpha.$$ Using $$a_1 = \frac{1}{x_1} = \alpha-\frac{\beta}{2},$$ and $$a_2 = \frac{1}{x_2} = \alpha+\frac{\beta}{4},$$ one gets $$\lim_{n\to\infty}x_n = \frac{1}{\alpha} = \frac{3x_1 x_2}{2x_1 + x_2}.$$

Answer 3

Solution

Since $$x_n = \frac{2}{\dfrac{1}{x_{n-1}} + \dfrac{1}{x_{n-2}}},$$ then $$ \frac{2}{x_{n}}=\frac{1}{x_{n-1}} + \frac{1}{x_{n-2}}.$$ Thus, $$2y_n=y_{n-1}+y_{n-2},$$ where $y_n=\dfrac{1}{x_n}$.

From $(1)$, we may obtain $$y_n-y_{n-1} = -\frac{1}{2}\left(y_{n-1}-y_{n-2}\right),$$ for $n=3,4,\cdots.$

Hence，$$y_n-y_{n-1}=(y_2-y_1)\left(-\frac{1}{2}\right)^{n-2}.$$

Thus, $$y_n-y_1=\sum_{k=2}^n(y_k-y_{k-1})=\sum_{k=2}^n(y_2-y_1)\left(-\frac{1}{2}\right)^{k-2}=(y_2-y_1)\cdot \frac{2}{3}\left[1-\left(-\frac{1}{2}\right)^{n-1}\right]$$

Thus, $$y_n=y_1+(y_2-y_1)\cdot \frac{2}{3}\left[1-\left(-\frac{1}{2}\right)^{n-1}\right].$$

Let $n \to \infty$. Then$$y_n \to y_1+(y_2-y_1)\cdot \frac{2}{3}=\frac{1}{3}y_1+\frac{2}{3}y_2=\frac{1}{3x_1}+\frac{2}{3x_2}=\frac{2x_1+x_2}{3x_1x_2}.$$

As a result, $$x_n=\frac{1}{y_n} \to \frac{3x_1x_2}{2x_1+x_2}.$$